Carga pontual em um campo elétrico

Carga pontual em um campo elétrico

• O que acontece com uma partícula carregada quando esta está na presença de um campo elétrico gerado por uma

distribuição de outras cargas elétricas estacionárias ou que se movem lentamente?

• A partícula fica sujeita a uma força eletrostática

F ~ = q E ~

Se q é positiva a força tem a direção do campo, do contrário, tem

direção oposta ao campo.

Aplicações  

(Impressora Jato de Tinta)

Idéia: Utilização de campos elétricos para controlar a impressão (movimento das gotículas de tinta)

Curiosidade: 100 gotas para formar uma letra

q < 0 F

G é um gerador que produz as gotículas de tinta

C é uma unidade de carregamento

que fornece carga às gotículas

Tubo de raios catódicos (TV’s antigas)

E

v₀ –

–

F I G U R E 2 1 - 2 6

Schematic drawing of a cathode-ray-tube display used for color television. The beams of electrons from the electron gun on the right activate phosphors on the screen at the left, giving rise to bright spots whose colors

depend on the relative intensity of each beam.

Electric fields between deflection plates in the gun (or magnetic fields from coils

surrounding the gun) deflect the beams. The beams sweep across the screen in a horizontal line, are deflected downward, then sweep across again. The entire screen is covered in this way 30 times per second. (Courtesy of Hulon Forrester/Video Display Corporation, Tucker Georgia.)

714 |

The Electric Field I: Discrete Charge Distributions

The convention relating the electric field strength to the density of the electric field lines works only because the electric field varies inversely as the square of the distance from a point charge. Because the gravitational field of a point mass also varies inversely as the square of the distance, field-line drawings are also useful for picturing gravitational fields. Near a point mass, the gravitational field lines ter- minate on the mass just as electric field lines terminate on a negative charge.

However, unlike electric field lines near a positive charge, there are no points in space from which gravitational field lines emanate. That is because the gravita- tional force between two masses is never repulsive.

21-6 ACTION OF THE ELECTRIC FIELD ON CHARGES

A uniform electric field can exert a force on a single charged particle and can exert both a torque and a net force on an electric dipole.

MOTION OF POINT CHARGES IN ELECTRIC FIELDS

When a particle that has a charge is placed in an electric field it experiences a force If the electric force is the only force acting on the particle, the particle has acceleration

where is the mass of the particle. (If the particle is an electron, its speed in an electric field is often a significant fraction of the speed of light. In such cases, Newton’s laws of motion must be modified by Einstein’s special theory of relativ- ity.) If the electric field is known, the charge-to-mass ratio of the particle can be de- termined from the measured acceleration. J. J. Thomson used the deflection of elec- trons in a uniform electric field in 1897 to demonstrate the existence of electrons and to measure their charge-to-mass ratio. Familiar examples of devices that rely on the motion of electrons in electric fields are oscilloscopes, computer monitors, and television sets that use cathode-ray-tube displays.

m

a

! © F

m ! q m E

q E

.

E

, q

Example 21-11 Electron Moving Parallel to a Uniform Electric Field

An electron is projected into a uniform electric field with an initial veloc- ity in the direction of the field (Figure 21-26). How far does the elec- tron travel before it is brought momentarily to rest?

PICTURE

Because the charge of the electron is negative, the force acting on the electron is in the direction opposite that of the field. Because is constant, the force is con- stant and we can use constant acceleration formulas from Chapter 2. We choose the field to be in the direction.

SOLVE

"x

E^S

F^S ! #eE^S v^S₀ !

(2.00

10

m

s)

(1000 N

C)

1. The displacement is related to the initial and final

velocities:

2a

2. The acceleration is obtained from Newton’s second law:

m ! #eE_x m

3. When

0, the displacement is:

1.14 cm

!

1.14

10

m

!

(9.11

10

kg)(2.00

10

m

s)

2(1.60

10

C)(1000 N

C)

¢x ! v²_x # v²_0x

2a

0

2(#eE

2eE

CHECK

The displacement

is positive, as is expected for something moving in the

direction.

Os feixes de elétrons emitidos pelo canhão de elétrons (direita) e ativam o material fluorescente da tela (esquerda) fazendo surgir

pontos brilhantes cujas cores dependem da intensidade de cada feixe.

Os feixes são desviados pelos campos elétricos entre as placas

defletoras do canhão.

Exemplos

1) Um elétron é projetado em um campo elétrico uniforme de módulo E=1000N/C na direção positiva do eixo x, com uma

velocidade inicial de 2x10 ⁶ m/s também na direção positiva do eixo x. Calcule a distância percorrida pelo elétron antes de atingir momentaneamente o repouso.

E

v₀ –

–

F I G U R E 2 1 - 2 6

Schematic drawing of a cathode-ray-tube display used for color television. The beams of electrons from the electron gun on the right activate phosphors on the screen at the left, giving rise to bright spots whose colors

depend on the relative intensity of each beam.

Electric fields between deflection plates in the gun (or magnetic fields from coils

surrounding the gun) deflect the beams. The beams sweep across the screen in a horizontal line, are deflected downward, then sweep across again. The entire screen is covered in this way 30 times per second. (Courtesy of

714 |

The Electric Field I: Discrete Charge Distributions

The convention relating the electric field strength to the density of the electric field lines works only because the electric field varies inversely as the square of the distance from a point charge. Because the gravitational field of a point mass also varies inversely as the square of the distance, field-line drawings are also useful for picturing gravitational fields. Near a point mass, the gravitational field lines ter- minate on the mass just as electric field lines terminate on a negative charge.

However, unlike electric field lines near a positive charge, there are no points in space from which gravitational field lines emanate. That is because the gravita- tional force between two masses is never repulsive.

21-6 ACTION OF THE ELECTRIC FIELD ON CHARGES

A uniform electric field can exert a force on a single charged particle and can exert both a torque and a net force on an electric dipole.

MOTION OF POINT CHARGES IN ELECTRIC FIELDS

When a particle that has a charge is placed in an electric field it experiences a force If the electric force is the only force acting on the particle, the particle has acceleration

where is the mass of the particle. (If the particle is an electron, its speed in an electric field is often a significant fraction of the speed of light. In such cases, Newton’s laws of motion must be modified by Einstein’s special theory of relativ- ity.) If the electric field is known, the charge-to-mass ratio of the particle can be de- termined from the measured acceleration. J. J. Thomson used the deflection of elec- trons in a uniform electric field in 1897 to demonstrate the existence of electrons and to measure their charge-to-mass ratio. Familiar examples of devices that rely on the motion of electrons in electric fields are oscilloscopes, computer monitors, and television sets that use cathode-ray-tube displays.

m

a

! © F

m ! q m E

q E

.

E

, q

Example 21-11 Electron Moving Parallel to a Uniform Electric Field

An electron is projected into a uniform electric field with an initial veloc- ity in the direction of the field (Figure 21-26). How far does the elec- tron travel before it is brought momentarily to rest?

PICTURE

Because the charge of the electron is negative, the force acting on the electron is in the direction opposite that of the field. Because is constant, the force is con- stant and we can use constant acceleration formulas from Chapter 2. We choose the field to be in the direction.

SOLVE

" x

E

F

! # eE

v

! (2.00 $ 10

m > s)i n E

! (1000 N > C) i n

1. The displacement is related to the initial and final

velocities: ¢ x v

! v

" 2a

¢ x

2. The acceleration is obtained from Newton’s second law: a

! F

m ! # eE

m

3. When v

! 0, the displacement is:

1.14 cm

! 1.14 $ 10

m !

! (9.11 $ 10

kg)(2.00 $ 10

m > s)

2(1.60 $ 10

C)(1000 N > C)

¢ x ! v

# v

2a

! 0 # v

2( # eE

> m) ! mv

2eE

CHECK

The displacement ¢ x is positive, as is expected for something moving in the " x direction.

Exemplos

2) Um elétron é projetado em um campo elétrico uniforme de módulo E=2000N/C na direção negativa do eixo y, com uma velocidade inicial de 10 ⁶ m/s na direção positiva do eixo x.

a) Compare as forças gravitacional e elétrica atuantes sobre este elétron.

b) Qual o valor do desvio sofrido pelo elétron após ter percorrido 1cm na direção x?

v₀ E –

–

F I G U R E 2 1 - 2 7

Action of the Electric Field on Charges S E C T I O N 2 1 - 6

|

CHECK The step-4 result is positive (upward), as is expected for an object accelerating upward that was initially moving horizontally.

TAKING IT FURTHER (a) As is usually the case, the electric force is huge compared with the gravitational force. Thus, it is not necessary to consider gravity when designing a cathode-ray tube, for example, or when calculating the deflection in the problem above.

In fact, a television picture tube works equally well upside down and right side up, as if gravity were not even present. (b) The path of an electron moving in a uniform electric field is a parabola, the same as the path of a neutral particle moving in a uniform gravi- tational field.

Example 21-12 Electron Moving Perpendicular to a Uniform Electric Field

An electron enters a uniform electric field with an initial velocity perpendicular to the field (Figure 21-27). (a) Compare the gravita- tional force acting on the electron to the electric force acting on it. (b) By how much has the electron been deflected after it has traveled 1.0 cm in the direction?

PICTURE (a) Calculate the ratio of the magnitude of the electric force to that of the gravitational force mg.(b) Because mgis, by comparison, negligible, the net force on the elec- tron is equal to the vertically upward electric force. The electron thus moves with constant horizontal velocity and is deflected upward by an amount where is the time to travel 1.0 cm in the direction.

SOLVE

x

¢y ! ¹₂at², t v_x

ƒ

ƒ

x

v^S₀ ! (1.0 " 10⁶ m>s)in E^S ! (#2.0 kN>C)jn

(a) 1. Calculate the ratio of the magnitude of the electric force, to the magnitude of the gravitational force, F_g:

F_e,

3.6 " 10¹³ F_e

F_g ! eE

mg ! (1.60 " 10^#19 C)(2000 N>C) (9.11 " 10^#31 kg)(9.81 N>kg) !

(b) 1. Express the vertical deflection in terms of the

acceleration and time t:a ¢y ! 1

2a_yt²

2. Express the time required for the electron to travel a horizontal distance with constant horizontal velocity v₀: ¢x

t ! ¢x v₀

3. Use this result for and t eE>mfor to calculate a_y ¢y:

1.8 cm

!

! 1 2

(1.6 " 10^#19 C)(2000 N>C)

9.11 " 10^#31 kg a0.010 m 10⁶ m>sb²

¢y ! 1 2

eE ma¢x

v₀ b²

Context-Rich

Example 21-13 The Electric Field in an Ink-Jet Printer

You have just finished printing out a long essay for your English professor, and you won- der about how the ink-jet printer knows where to place the ink. You search the Internet and find a picture (Figure 21-28) showing that the ink drops are given a charge and pass between a pair of oppositely charged metal plates that provide a uniform electric field in the region between the plates. Because you have been studying the electric field in physics class, you wonder if you can determine how large a field is used in this type of printer. You search further and find that the ink drops have an initial velocity of and that a drop that has a 2.00-nC charge is deflected upward a dis- tance of 3.00 mm as the drop travels through the 1.00-cm-long region between the plates.

Find the magnitude of the electric field. (Neglect any effects of gravity on the motion of the drops.)

40.0 m>s,

40.0-mm-diameter

Exercícios Lista IV

Campos Elétricos  

Bibliografia, Figuras e Exemplos: Halliday, Resnick e Walker, 8a ed. vol3, capítulo 22   Tipler e Mosca, vol 2, 6a edição.  

• Nas aulas anteriores aprendemos a calcular o vetor campo elétrico devido à uma distribuição discreta de cargas, envolvendo poucas cargas elétricas.

• Como o campo elétrico é uma grandeza vetorial vimos que, assim como no caso da força eletrostática, vale o princípio de superposição, ou seja, o campo elétrico

resultante em uma carga de prova q 0 devido a um sistema de n-cargas é calculado através de:

E ~ _R =

X n

i=1

k q _i

r _i ² r ˆ

Campo Elétrico devido à uma distribuição contínua de cargas

• Quando o número de cargas que geram os campos elétricos é muito grande, dizemos que temos uma distribuição contínua de cargas elétricas.

• Aprenderemos a calcular o campo elétrico resultante destas distribuições de carga quando:

• as cargas estiverem distribuídas ao longo de uma linha

• as cargas estiverem distribuídas ao longo de uma superfície

• as cargas estiverem distribuídas ao longo de um volume

Campos elétricos devidos à distribuições contínuas de cargas

• Quando falamos em distribuições contínuas de cargas é melhor trabalharmos com as densidades de carga.

• No caso de uma linha de cargas temos a densidade linear de carga, , cuja unidade é o C/m.

• No caso de uma superfície carregada trabalhamos com a densidade superficial de carga, , cuja unidade no SI é o C/m ²

• No caso de um volume carregado usaremos a densidade volumétrica de carga, , cuja unidade no SI é o C/m ⇢ ³

E ~ =

Z

d E ~ =

Z

k dq

r ² r ˆ

Campo elétrico sobre o eixo de um segmento de reta finito.

++++++++ ++++++++

x

x

x

dx

x x

-x P

E

dq = dx ^{x 1 =}

L 2 x ₂ = L

2

dE _x ˆ i = k dq

(x _p x) ² ˆ i

O campo elétrico devido ao elemento de carga dq em P é:

dE _x ˆ i = k dx

(x _p x) ² ˆ i

Para o segmento todo temos que integrar esta expressão entre x 1 e

x 2

Campo elétrico sobre o eixo de um segmento de reta finito.

E _x = k

Z _L/2

L/2

dx

(x _p x) ²

Como resolver a integral? u = x _p x ! du = dx

chegando facilmente (exercício) em

E _x = k

Z _{x p L/2}

x _p +L/2

du u ²

= k

Z x _p L/2 x _p +L/2

u ² du

E ~ = k L

x ² _p (L/2) ² ˆı

Campo elétrico sobre o eixo de um segmento de reta finito.

q = L

Note que é a carga total do segmento de reta.

O que acontece com o campo elétrico para x p >> L?

Linhas de Carga  

1) Anel de raio R carregado positivamente.

= dq

ds ! dq = ds

Este elemento de carga dq produz um campo elétrico dE no ponto P.

dE = k dq r ²

= k ds

r ² da figura vemos que r ² =z ² +R ²

dE = k ds

z ² + R ²

Linhas de Carga  

1) Continuação: Anel de raio R carregado positivamente.

Note que dE possui componentes vertical e horizontal. Por simetria, a cada componente horizontal existe outra igual e

oposta que a anula. Ficamos então somente com as componentes verticais que vão ser somadas. Da figura vemos facilmente que:

E = Z

dEcos✓

cos✓ = z

r = z

(z ² + R ² ) ^1/2

E =

Z k z

(z ² + R ² ) ^3/2 ds

= k z

(z ² + R ² ) ^3/2

Z 2⇡R 0

ds E = kz (2⇡R)

(z ² + R ² ) ^3/2

Como q = (2⇡ R) temos que:

E = kzq

(z ² + R ² ) ^3/2

Note que

Se z = 0, E = 0 Se z R, E = kq

z ²

Campo elétrico produzido por um disco carregado em um ponto P sobre o eixo z.

= dq

dA ! dq = dA = (2⇡ rdr)

Note que já resolvemos o problema de um anel carregado e o resultado encontrado foi:

dE = kzdq

(z ² + R ² ) ^3/2

Fazendo R = r e substituindo dq ficamos com:

dE = kz 2⇡ rdr

(z ² + r ² ) ^3/2 ! k = 1 4⇡✏ ₀

= z

4✏ ₀

2rdr

(z ² + r ² ) ^3/2

Para calcular o campo elétrico total, devemos integrar o resultado anterior na variável dr, entre 0 e R, desta maneira estamos

“somando” os vários anéis de carga, preenchendo a área do disco carregado, ou seja:

E =

Z _R

0

dE

=

Z _R

0

z 4✏ ₀

2rdr

(z ² + r ² ) ^3/2

= z

4✏ ₀

Z _R

0

2rdr

(z ² + r ² ) ^3/2

Como resolver esta integral? (Exercício) u = z ² + r ^{2 du = 2rdr}

E = z 4✏ ₀

Z z ² +R ² z ²

du

u ^3/2 E = z

4✏ ₀ [ u ^1/2

1 2

] ^z _z ² 2 ^{+R 2}

Substituindo os limites de integração e fazendo um pouco de álgebra (exercício) chegamos à

E =

2✏ ₀ (1 z

p z ² + R ² )

Se z mantido finito e, R ! 1 , E =

2✏ ₀